Question

If sin A = cos B then prove that \[A + B = {90^ \circ }\]

Answer 1

Hint: Take any value from \[{0^ \circ }\] to \[{90^ \circ }\] for A in \[\sin A\] and then see for what possible value of B we can get \[\sin A = \cos B\] then add A and B to get the required results.
Complete Step by Step Solution:
Let us do it by taking some examples, we know that \[\sin \dfrac{\pi }{6} = \dfrac{1}{2}\& \cos \dfrac{\pi }{3} = \dfrac{1}{2}\]
So here we can say that \[\sin \dfrac{\pi }{6} = \cos \dfrac{\pi }{3} = \dfrac{1}{2}\]
Now Here i am taking \[A = \dfrac{\pi }{6}\& B = \dfrac{\pi }{3}\]
So after adding them both we will get it as
\[\begin{array}{l}
\therefore \dfrac{\pi }{6} + \dfrac{\pi }{3}\\
 = \dfrac{{\pi + 2\pi }}{6}\\
 = \dfrac{{3\pi }}{6}\\
 = \dfrac{\pi }{2}
\end{array}\]
It is also known that \[\dfrac{\pi }{2} = {90^ \circ }\]
The same goes for \[\dfrac{\pi }{4}\]
Here in this case \[A = B = \dfrac{\pi }{4}\]
And again \[A + B = \dfrac{\pi }{4} + \dfrac{\pi }{4} = \dfrac{\pi }{2}\]

Note: we also know that \[\sin A = \cos (90 - A)\] so this also settles it because we can also write \[\cos B = \sin (90 - B)\] therefore \[\sin A = \sin (90 - B)\] and now if we remove sin from both the sides we are left with \[A = {90^ \circ } - B\] i.e., \[A + B = {90^ \circ }\] .